Recent content by Lacnez
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Harmonic motion and acceleration
don't i lack the required information to use a motion equation- Lacnez
- Post #3
- Forum: Introductory Physics Homework Help
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Harmonic motion and acceleration
A system exhibits simple harmonic motion with a frequency of 0.85 cycles per second. Calculate the acceleration experienced by the mass 3.0 m from the equilibrium This question seems simple, I haven't really tried anything cause I can't figure out how to start. I need help to start this...- Lacnez
- Thread
- Acceleration Harmonic Harmonic motion Motion
- Replies: 4
- Forum: Introductory Physics Homework Help
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Calculate Momentum Change for a 1.0 kg Firecracker at 25 m/s Launch Speed
alright thanks a lot guy I apologize for the inconvenience I'm doing this course all online and no one to help or teach it to me so it's not the greatest thing. Thanks again- Lacnez
- Post #34
- Forum: Introductory Physics Homework Help
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Calculate Momentum Change for a 1.0 kg Firecracker at 25 m/s Launch Speed
alright i did 12.5tan7.5 = 1.6457 kg m/s- Lacnez
- Post #32
- Forum: Introductory Physics Homework Help
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Calculate Momentum Change for a 1.0 kg Firecracker at 25 m/s Launch Speed
ok so i leave it as is at 1.6315 or do i add 1.6315 + 1.6315 to get the final answer- Lacnez
- Post #30
- Forum: Introductory Physics Homework Help
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Calculate Momentum Change for a 1.0 kg Firecracker at 25 m/s Launch Speed
so the answer to my question of change in momentum would be the 3.263- Lacnez
- Post #27
- Forum: Introductory Physics Homework Help
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Calculate Momentum Change for a 1.0 kg Firecracker at 25 m/s Launch Speed
The horizontal before the split was 0 cause it was going vertical For one of the pieces I did now P=mv (0.5)(25) =12.5 Then 12.5sin7.5 I get as a horizontal 1.6315 Then subtract 1.6315-(-1.6315) = 3.263- Lacnez
- Post #25
- Forum: Introductory Physics Homework Help
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Calculate Momentum Change for a 1.0 kg Firecracker at 25 m/s Launch Speed
i took 25sin7.5 got 3.263 the subtracted 3.263 - (-3.263) and got 6.52 as an overall horizontal- Lacnez
- Post #23
- Forum: Introductory Physics Homework Help
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Calculate Momentum Change for a 1.0 kg Firecracker at 25 m/s Launch Speed
would it after the split subtracting the horizontal vectors give you some were around 6.52k kg m/s- Lacnez
- Post #21
- Forum: Introductory Physics Homework Help
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Calculate Momentum Change for a 1.0 kg Firecracker at 25 m/s Launch Speed
im assuming you also get 0 for the horizontal- Lacnez
- Post #19
- Forum: Introductory Physics Homework Help
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Calculate Momentum Change for a 1.0 kg Firecracker at 25 m/s Launch Speed
when you subtract the verticals you get 0- Lacnez
- Post #17
- Forum: Introductory Physics Homework Help
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Calculate Momentum Change for a 1.0 kg Firecracker at 25 m/s Launch Speed
I just don't know, what I'm thinking you're saying is to original firecracker VERTICAL 25kg m/s HORIZONTAL 0 kg m/s One of the pieces VERTICAL 25 kg m/s HORIZONTAL 3.263 kg m/s and then subtract the components but that doesn't make sense to me- Lacnez
- Post #15
- Forum: Introductory Physics Homework Help
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Calculate Momentum Change for a 1.0 kg Firecracker at 25 m/s Launch Speed
the change is 7.5 degrees either direction- Lacnez
- Post #13
- Forum: Introductory Physics Homework Help
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Calculate Momentum Change for a 1.0 kg Firecracker at 25 m/s Launch Speed
i believe that there is not change or very slight change in vertical velocities after the separation compared to before separation- Lacnez
- Post #11
- Forum: Introductory Physics Homework Help
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Calculate Momentum Change for a 1.0 kg Firecracker at 25 m/s Launch Speed
mass of one part would be half of the original mass so 0.5 kg is the relationship the change in vertical velocities before and after separation and is the triangle representing the change of direction of momentum- Lacnez
- Post #9
- Forum: Introductory Physics Homework Help